Group Theory

1. Show that the set {0} with addition is a group.2. Show that the set {0} with multiplication is a group.3. Show that the set {1} with addition is not a group.4. Show that the set {1} with multiplication is a group.5. Show that the set {-1, 1} is a group under multiplication, but not addition.6. Name the multiplicative inverse for -1 in the group {-1, 1} under multiplication.7. Show that if abab = aabb, then it must be that ab = ba.8. Show that the matrix [1 2][2 1] * [3 2][1 2] does not equal [3 2][1 2] * [1 2][2 1] (It may be noted that matrices of integers are groups). This would mean that matrices are not abelian.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

Re: Group Theory

3. Show that the set {1} with addition is not a group.For any elements a and b of {1}, (a+b) is not an element of {1}. 1+1=2. It is enough to show only one rule-break to prove that {1} is not a group with respect to addition.We conclude {1} is not a group with respect to addition.

Re: Group Theory

5. Show that the set {-1, 1} is a group under multiplication, but not addition.For any elements a and b of {-1, 1}, (a+b) is not an element of {-1, 1}. It is enough to show only one rule-break to prove that {-1, 1} is not a group with respect to addition.We conclude {-1, 1} is not a group with respect to addition.------------------------------------------------------------------------------------------------------------------------For any elements a and b of {-1,1}, (a*b) is an element of {1,-1}. The closure law has been followed.

For any a,b,c of {1,-1}; a*(b*c) = (a*b)*c. The associative law has been followed.

For any a of {1,-1} i*a=a, where i is a particular element in {1,-1}.The left identity element i is 1 here.

For any a of {1,-1} the equation x*a=i has a solution known as the left inverse of a.

All these properties are followed by this set that is closed under multiplication.Therefore, {1,-1} is a group with respect to multiplication.

Re: Group Theory

I don't think they are a group under multiplication, because not all the matrices in the set have inverses...

Here lies the reader who will never open this book. He is forever dead.Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and PunishmentThe knowledge of some things as a function of age is a delta function.

Re: Group Theory

Great!!

Why didn't I see that before?

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.''God exists because Mathematics is consistent, and the devil exists because we cannot prove it''You have made another human being happy. There is no greater accomplishment.' -bobbym

Re: Group Theory

What is abstract algebra like?

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.''God exists because Mathematics is consistent, and the devil exists because we cannot prove it''You have made another human being happy. There is no greater accomplishment.' -bobbym

Re: Group Theory

It's pretty abstract and broad. You start learning about algebraic structures by them selves, like groups, rings, fields etc. You start learning why things actually work (certain properties). I preferred analysis over it for quite a bit, but I'm starting to like it more nowadays.

Re: Group Theory

The greatest tool in the mathematician's tool chest is abstraction. I cannot count all the times when I've done some long, horrific calculation or proof, and some time later came across an abstraction that made the whole thing almost trivial.

Example: prove that det(AB) = det(A)det(B). If you try to do that for higher dimensions by looking at the coordinate formula, you'll be gibbering before very long (2D isn't bad, 3 is a pain, 4 is awful, much higher, and you might as well throw the whole thing in). But if you develop the concept of vector spaces, and introduce the wedge product, then suddenly, the determinant pops up in such a way that the det(AB) = det(A)det(B) result is a trivial consequence.

I find analysis more interesting myself, but abstract algebra shows up in pretty much all other fields, so it is a very good thing to master.

"Having thus refreshed ourselves in the oasis of a proof, we now turn again into the desert of definitions." - Bröcker & Jänich